A) parallel to position vector
B) perpendicular to position vector
C) directed towards the origin
D) directed away from the origin
Correct Answer: B
Solution :
Position vector \[\overset{\to }{\mathop{r}}\,=(a\cos \omega t)\widehat{\text{i}}+(a\sin \omega t)\widehat{\text{j}}\] \[\therefore \]velocity\[\vec{v}=\frac{d\vec{r}}{dt}\] \[=\frac{d}{dt}[(a\cos \omega t)\text{\hat{i}}+(a\sin \omega t)\text{\hat{j}}]\] \[=(-a\sin \omega t)\text{\hat{i}}+(a\cos \omega \,\,t)\text{\hat{j}}\] \[\therefore \] \[\mathbf{\vec{v}}\cdot \mathbf{\vec{r}}=[(-a\sin \omega t)\text{\hat{i}}+(a\cos \omega t)\text{\hat{j}}]\] \[[(a\cos \omega t)\text{\hat{i}}+(a\sin \omega t)\text{\hat{j}}]\] \[=-{{a}^{2}}\cos \omega t\sin \omega t+{{a}^{2}}\omega t\cos \omega t=0\] \[\therefore \] \[\vec{v}\bot \vec{r}\]i.e., velocity vector is perpendicular to position vector.
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